Let $F$ be a field. A field containing $F$ is called a field extensions of $F$ or an extension of $F$.

When considering field extensions of $F$, the field $F$ is sometimes called the base field.

• Every field of characteristic $0$ is a field extension of $\mathbb{Q}$.
• Every field of characteristic $p$ is a field extension of $\mathbb{F}_{p}$.

1. $\mathbb{C}$ is a field extension of $\mathbb{R}$ and $[\mathbb{C} : \mathbb{R}] = 2$.
2. $\mathbb{R}$ is a field extension of $\mathbb{Q}$ and $[\mathbb{R} : \mathbb{Q}] = \infty$.
3. Let $F$ be a field and $p(x)$ be an irreducible polynomial of degree $\geq 1$ in $F[x]$. Then $K = F[x] / (p(x))$ is a field extension of $F$ and $[K : F] = \deg p(x)$.
4. $F = \mathbb{Q}, p(x) = x^{2} - 2, K = \mathbb{Q}[x] / (x^{2} - 2)$. Then $K \cong \mathbb{Q}[\sqrt{ 2 }] = \{ a + b\sqrt{ 2 } \mid a, b \in \mathbb{Q} \}$.
5. $[\mathbb{F}_{p}(t), \mathbb{F}_{p}] = \infty$. $1, t, t^{2}, \dots, t^{n}, \dots$ are linearly independent over $\mathbb{F}_{p}$, so $\dim_{\mathbb{F}_{p}}\mathbb{F}_{p}(t) = \infty$.

Let $F$ be a field and $p(x)$ be an irreducible polynomial in $F[x]$. Then

1. $K = F[x] / (p(x))$ is a field extension of $F$,
2. $\theta = \overline{x} \in K$ is a root of $p(x)$ in $K$,
3. $[K : F] = \deg p(x)$. Let $n = \deg p(x)$. Then $$1, \theta, \theta^{2}, \dots, \theta^{n-1}$$ is a basis of $K$ as an $F$-vector space. (Hence, $K = \{ a_{0} + a_{1} \theta + \cdots + a_{n-1}\theta^{n-1} \mid a_{0}, \dots, a_{n-1} \in F\}$.)

Let $F \subseteq K \subseteq L$ be fields. Then $[L:F]$ is finite if and only if $[L:K]$ and $[K:F]$ are both finite, and $$ [L:F] = [L : K][K : F]. $$

When $[L : F] = \infty$, at most one of $[L:K]$ and $[K:F]$ can be finite. For example, $[\mathbb{C} : \mathbb{Q}] = \infty$ and $[\mathbb{R} : \mathbb{Q}] = \infty$ while $[\mathbb{C} : \mathbb{R}] = 2 < \infty$.

Suppose $f_{n} \to f$ uniformly on a set $E$ in a metric space. Let $x$ be a limit point of $E$, and suppose that $$ \lim_{ t \to x } f_{n}(t) = A_{n} \quad (n = 1, 2, 3, \dots). $$ Then $\{ A_{n} \}$ converges, and $$ \lim_{ t \to x } f(t) = \lim_{ n \to \infty } A_{n}. $$ In other words, the conclusion is that $$ \lim_{ t \to x } \lim_{ n \to \infty } f_{n}(t) = \lim_{ n \to \infty } \lim_{ t \to x } f_{n}(t). $$

If $\{ f_{n} \}$ is a sequence of continuous functions on $E$, and if $f_{n} \to f$ uniformly on $E$, then $f$ is continuous on $E$.

A family $\mathscr{F}$ of complex functions $f$ defined on a set $E$ in a metric space $X$ is said to be equicontinuous on $E$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $$ \lvert f(x) - f(y) \rvert < \epsilon $$ whenever $d(x, y) < \delta, x \in E, y \in E$, and $f \in \mathscr{F}$. Here $d$ denotes the metric of $X$.

It is clear that every member of an equicontinuous family is uniformly continuous.

If $\{ f_{n} \}$ is a pointwise bounded sequence of complex functions on a countable set $E$, then $\{ f_{n} \}$ has a subsequence $\{ f_{n_{k}} \}$ such that $\{ f_{n_{k}}(x) \}$ converges for every $x \in E$.

If $K$ is a compact metric space, if $f_{n} \in \mathscr{C}(K)$ for $n = 1, 2, 3, \dots$, and if $\{ f_{n} \}$ converges uniformly on $K$, then $\{ f_{n} \}$ is equicontinuous on $K$.

If $K$ is compact, if $f_{n} \in \mathscr{C}(K)$ for $n = 1, 2, 3, \dots$, and if $\{ f_{n} \}$ is pointwise bounded and equicontinuous on $K$, then

1. $\{ f_{n} \}$ is uniformly bounded on $K$,
2. $\{ f_{n} \}$ contains a uniformly convergent subsequence.

A mapping $A$ of a vector space $X$ into a vector space $Y$ is said to be a linear transformation if $$ A (\mathbf{x}_{1} + \mathbf{x}_{2}) = A \mathbf{x}_{1} + A \mathbf{x}_{2}, \quad A(c \mathbf{x}) = cA \mathbf{x} $$ for all $\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x} \in X$ and all scalars $c$.

1. Let $L(X, Y)$ be the set of all linear transformations of the vector space $X$ into the vector space $Y$. Instead of $L(X, X)$, we shall simply write $L(X)$.

2. If $X, Y, Z$ are vector spaces, and if $A \in L(X, Y)$ and $B \in L(Y,Z)$, we define their product $BA$ to be the composition of $A$ and $B$: $$(BA)\mathbf{x} = B(A\mathbf{x}) \qquad (\mathbf{x} \in X).$$ Then $BA \in L(X, Z)$.

3. For $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, define the norm $\lVert A \rVert$ of $A$ to be the sup of all numbers $\lvert A\mathbf{x} \rvert$, where $\mathbf{x}$ ranges over all vectors in $\mathbb{R}^{n}$ with $\lvert \mathbf{x} \rvert \leq 1$.

   Observe that the inequality $$\lvert A\mathbf{x} \rvert \leq \lVert A \rVert \lvert \mathbf{x} \rvert $$ holds for all $\mathbf{x} \in \mathbb{R}^{n}$. Also, if $\lambda$ is such that $\lvert A\mathbf{x} \rvert \leq \lambda \lvert \mathbf{x} \rvert$ for all $\mathbf{x} \in \mathbb{R}^{n}$, then $\lVert A \rVert \leq\lambda$.

1. If $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, then $\lVert A \rVert < \infty$ and $A$ is a uniformly continuous mapping of $\mathbb{R}^{n}$ into $\mathbb{R}^{m}$.
2. If $A, B \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$ and $c$ is a scalar, then$$\lVert A + B \rVert \leq \lVert A \rVert + \lVert B \rVert , \qquad \lVert cA \rVert = \lvert c \rvert \lVert A \rVert.$$With the distance between $A$ and $B$ defined as $\lVert A - B \rVert$, $L(\mathbb{R}^{n}, \mathbb{R}^{m})$ is a metric space.
3. If $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$ and $B \in L(\mathbb{R}^{m}, \mathbb{R}^{n})$, then$$\lVert BA \rVert \leq \lVert B \rVert \lVert A \rVert.$$

   Proof. sketch [br97A]

   1. Let $\{ \mathbf{e}_{1}, \dots, \mathbf{e}_{n} \}$ be the standard basis in $\mathbb{R}^{n}$ and suppose $\mathbf{x} = \sum c_{i}\mathbf{e}_{i}$. Now $\lvert \mathbf{x} \rvert \leq 1 \implies \lvert c_{i} \rvert \leq 1$ for all $i$. Then $$\lvert A \mathbf{x} \rvert = \left\lvert \sum c_{i} A \mathbf{e}_{i} \right\rvert \leq \sum \lvert c_{i} \rvert \lvert A \mathbf{e}_{i} \rvert \leq \sum \lvert A \mathbf{e}_{i} \rvert$$ so that $$\lVert A \rVert \leq \sum_{i=1}^{n} \lvert A \mathbf{e}_{i} \rvert < \infty. $$

      Each $A \mathbf{e}_{i}$ is a specific vector in $\mathbb{R}^{m}$, and hence $\lvert A \mathbf{e}_{i} \rvert < \infty$.

      Since $\lvert A \mathbf{x} - A \mathbf{y} \rvert \leq \lVert A \rVert \lvert \mathbf{x} - \mathbf{y} \rvert$ if $\mathbf{x}, \mathbf{y} \in \mathbb{R}^{n}$, we see that $A$ is uniformly continuous.

   2. Omitted.

   3. Omitted.

Let $\Omega$ be the set of all invertible linear operators on $\mathbb{R}^{n}$.

1. If $A \in \Omega, B \in L(\mathbb{R}^{n})$, and$$\lVert B - A \rVert \cdot \lVert A^{-1} \rVert < 1,$$then $B \in \Omega$.
2. $\Omega$ is an open subset of $L(\mathbb{R}^{n})$, and the mapping $A \to A^{-1}$ is continuous on $\Omega$.

Suppose $\{ \mathbf{x}_{1}, \dots, \mathbf{x}_{n} \}$ and $\{ \mathbf{y}_{1}, \dots, \mathbf{y}_{m} \}$ are bases of vector spaces $X$ and $Y$, respectively. Then every $A \in L(X, Y)$ determines a set of numbers $a_{ij}$ such that $$ A \mathbf{x}_{j} = \sum_{i=1}^{m} a_{ij} \mathbf{y}_{i} \qquad (1 \leq j \leq n). $$ We represent $A$ by an $m$ by $n$ matrix: $$ [A] = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{mn} \end{bmatrix} $$

$[A]$ depends not only on $A$ but also on the choice of bases in $X$ and $Y$.

If $\mathbf{x} = \sum_{j} \mathbf{x}_{j}$, then the Schwarz inequality shows that $$ \lvert A \mathbf{x} \rvert^{2} = \sum_{i} \left( \sum_{j} a_{ij} c_{j} \right)^{2} \leq \sum_{i} \left( \sum_{j} a_{ij}^{2} \cdot \sum_{j} c_{j}^{2} \right) = \sum_{i, j} a_{ij}^{2} \lvert \mathbf{x} \rvert ^{2}. $$ Thus $$ \lVert A \rVert \leq \sqrt{ \sum_{i, j} a_{ij}^{2} }. $$ Moreover, if we replace $A$ by $B - A$, where $A, B \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, and view each $a_{ij}$ as continuous functions of a single parameter, then we have the following:

If $S$ is a metric space, if $a_{11}, \dots, a_{mn}$ are real continuous functions on $S$, and if, for each $p \in S$, $A_{p}$ is the linear transformation of $\mathbb{R}^{n}$ into $\mathbb{R}^{m}$ whose matrix has entries $a_{ij}(p)$, then the mapping $p \mapsto A_{p}$ is a continuous mapping of $S$ into $L(\mathbb{R}^{n}, \mathbb{R}^{m})$.